A convex quadrilateral $ABCD$ is inscribed and circumscribed. If the diagonals $AC$ and $BD$ are perpendicular, show that one of them divides the quadrilateral into two congruent right triangles.
The first question I asked myself is how we determine which of the diagonals divides $ABCD$ in the desired way.
If $AB=a,BC=b,CD=c,AD=d$, then $a+c=b+d$ on one hand. Also we have $A+C=B+D=180^\circ$. I tried to do something with the areas. $$S_{ABC}=\dfrac12ab\sin B\\S_{ACD}=\dfrac12cd\sin D\\S_{ABCD}=\dfrac12ab\sin B+\dfrac12cd\sin D=\dfrac12\sin B(ab+cd)\\S_{ABCD}=\dfrac{AC\cdot BD}{2}$$ I don't really think anything of that helps. How to approach the problem?
Best Answer
Let the diagonals intersect at $O$ and set: $a=AO$, $b=BO$, $c=CO$, $d=DO$.
The sums of opposite sides must be the same, giving: $$ \strut\sqrt{a^2+b^2}+\sqrt{c^2+d^2}= \sqrt{a^2+d^2}+\sqrt{b^2+c^2}. $$ Squaring and simplifying we get: $$ (a^2+b^2)(c^2+d^2)= (a^2+d^2)(b^2+c^2), $$ which reduces to $$ (a^2-c^2)(b^2-d^2)=0. $$ Hence, $a=c$ or $b=d$, that is one of the diagonals bisects the other one and is thus a diameter of the circumscribed circle.